\newproblem{lay:6_7_13}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 6.7.13}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Let $A$ by any invertible $n\times n$ matrix. Show that for $\mathbf{u}$ and $\mathbf{v}$ in $\mathbb{R}^n$, the formula
	$\left<\mathbf{u},\mathbf{v}\right>=(A\mathbf{u})^T(A\mathbf{v})$ defines an inner product in $\mathbb{R}^n$.
}{
   % Solution
	To show that the proposed operation is an inner product we need to show all the properties below:
	\begin{enumerate}[a.]
		\item $\left<\mathbf{u},\mathbf{v}\right>=\left<\mathbf{v},\mathbf{u}\right>$
					\begin{center}
						$\begin{array}{rcll}
							\left<\mathbf{u},\mathbf{v}\right>&=&(A\mathbf{u})^T(A\mathbf{v}) & \text{[by definition]} \\
							   &=&\mathbf{u}^TA^TA\mathbf{v} & \\
							   &=&(\mathbf{u}^TA^TA\mathbf{v})^T& \text{the result of the inner product is a scalar} \\
							   &=&\mathbf{v}^TA^TA\mathbf{u}& \\
							   &=&(A\mathbf{v})^T(A\mathbf{u})& \\
							   &=&\left<\mathbf{v},\mathbf{u}\right>& \\
						\end{array}$
					\end{center}
		\item $\left<\mathbf{u}+\mathbf{v},\mathbf{w}\right>=\left<\mathbf{u},\mathbf{w}\right>+\left<\mathbf{v},\mathbf{w}\right>$
					\begin{center}
						$\begin{array}{rcll}
							\left<\mathbf{u}+\mathbf{v},\mathbf{w}\right>&=&(A(\mathbf{u}+\mathbf{v}))^T(A\mathbf{w}) & \text{[by definition]} \\
							   &=& (A\mathbf{u}+A\mathbf{v})^T(A\mathbf{w}) & \\
							   &=& (\mathbf{u}^TA^T+\mathbf{v}^TA^T)(A\mathbf{w}) & \\
							   &=& \mathbf{u}^TA^TA\mathbf{w}+\mathbf{v}^TA^TA\mathbf{w} & \\
							   &=& (A\mathbf{u})^T(A\mathbf{w})+(A\mathbf{v})^T(A\mathbf{w}) & \\
							   &=& \left<\mathbf{u},\mathbf{w}\right>+\left<\mathbf{v},\mathbf{w}\right> & \\
						\end{array}$
					\end{center}
		\item $\left<c\mathbf{u},\mathbf{v}\right>=c\left<\mathbf{u},\mathbf{v}\right>$
					\begin{center}
						$\begin{array}{rcll}
							\left<c\mathbf{u},\mathbf{v}\right>&=&(A(c\mathbf{u}))^T(A\mathbf{v}) & \text{[by definition]} \\
							   &=&(cA\mathbf{u})^T(A\mathbf{v}) & \\
							   &=&c\mathbf{u}^TA^TA\mathbf{v} & \\
							   &=&c(A\mathbf{u})^T(A\mathbf{v}) & \\
							   &=&c\left<\mathbf{u},\mathbf{v}\right> & \\
						\end{array}$
					\end{center}
		\item $\left<\mathbf{u},\mathbf{u}\right> \geq 0 $.
					\begin{center}
						$\begin{array}{rcll}
							\left<\mathbf{u},\mathbf{u}\right>&=&(A\mathbf{u})^T(A\mathbf{u}) & \text{[by definition]} \\
							   &=&\|A\mathbf{u}\|^2\geq 0 & \\
						\end{array}$
					\end{center}
		\item $\left<\mathbf{u},\mathbf{u}\right>=0$ iff $\mathbf{u}=\mathbf{0}$.
					\begin{center}
						$\left<\mathbf{u},\mathbf{u}\right>=0\Rightarrow \|A\mathbf{u}\|^2=0 \Rightarrow A\mathbf{u}=\mathbf{0}$
					\end{center}
					According to the requirement of inner products, it must be that
					\begin{center}
						$A\mathbf{u}=\mathbf{0} \Leftrightarrow \mathbf{u}=\mathbf{0}$
					\end{center}
					This means that it must be $\mathrm{Nul}\{A\}=\{\mathbf{0}\}$. For an $n\times n$ matrix, this only happens if and only if
					$A$ is invertible (as stated by the problem; see the Invertible Matrix Theorem).
	\end{enumerate}
	Since the proposed inner product meets all the conditions, it is a true inner product.
}
\useproblem{lay:6_7_13}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}

